Question: We know that $ \dfrac{n^2}{n^3-\ln(n)} \ge \dfrac{n^2}{n^3}=\dfrac{1}{n}>0$ for any $n\ge 1$. Considering this fact, what does the direct comparison test say about $\sum\limits_{n=1}^{\infty }~\dfrac{n^2}{n^3-\ln(n)}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A The series converges. (Choice B) B The series diverges. (Choice C) C The test is inconclusive.
Answer: $\sum_{n=1}^{\infty }{\frac{1}{{n}}}~$ is the harmonic series which is known to diverge. Our given series is term-by-term greater than or equal to a divergent series, so it also diverges.